3
$\begingroup$

Let $\big( X_t , t\geq 0 \big)$ be a measurable process, that is, $$\big( t, \omega \big) \in \mathbb{R}_+\times\Omega\longmapsto X_t(\omega)\in\mathbb{R} \quad\text{is $\mathscr{B}\big( \mathbb{R}_+ \big)\otimes\mathscr{F} \big/\mathscr{B}\big(\mathbb{R} \big)$-measurable. } $$ $T : \Omega\to \big[ 0, +\infty \big]$ be a random time.

Show that the collection $\mathscr{G}$ of all sets of the form $\big( X_T\in A \big)$ and $\big( X_T\in A \big)\cup \big( T = \infty \big)$, $A\in\mathscr{B}(\mathbb{R} )$ is a $\sigma$-algebra on $\Omega$.

Problem 1.17. Karatzas et Shreve.

My confusion is how to understand this question given that $X_T$ is undefined on $\big( T = +\infty \big)$.


EDIT: there would be no problem if we make a change as follows:

Show that the collection $\mathscr{G}$ of all sets of the form $\big( X_T\in A , T < +\infty \big)$ and $\big( X_T\in A , T < +\infty\big)\cup \big( T = \infty \big)$, $A\in\mathscr{B}(\mathbb{R} )$ is a $\sigma$-algebra on $\Omega$.

$\endgroup$
1
  • $\begingroup$ $\big( X_T\in A , T < +\infty \big)=\big( X_T\in A \big)$, so there is no undefined question in this problem. $\endgroup$ Commented May 12, 2015 at 13:06

1 Answer 1

4
$\begingroup$

One way of defining $X_T$ on the whole of $\Omega$ is to let $$ X_T(\omega):= \begin{cases} \lim\limits_{n\to\infty} X_{T(\omega)\wedge n}(\omega)\quad &\text{if the limits exists and is finite},\\ 0 &\text{otherwise}. \end{cases} $$ With this definition we have a random variable $X_T$ defined on $\Omega$ with $X_T(\omega)=X_{T(\omega)}(\omega)$ for $\omega\in\{T<\infty\}$.

$\endgroup$

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.